Use the product of the generating functions g(x, t) and g(x,−t) to show that 1 = [J0(x)]2+ 2∑n=1[Jn(x)]^2. (Hint: a sneaky use of the residue theorem will work wonders here.

For part (a), note that you also have

g(x,−t)=g(x,1/t)=∑n=0∞Jn(x)(1t)n.

You can then multiply this with the original series to get

[∑n=−∞∞Jn(x)t−n][∑n=−∞∞Jn(x)tn]=1.

By multiplying these two series together and matching coefficients, you should be able to prove the given identity.

Parts (b) and (c) follow from the identity in part (a). Note that if you have a set of quantities ai such that ai>0 for all i, then ai<∑jaj for all i.

Step-by-step explanation:

4x and 6y

im not sure though

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all